To evaluate the double integral \(\iint_{D} (x^2 + 6y) \, dA\), we first need to identify the region D defined by the curves \(y = x\) and \(y = x^3\) in the first quadrant (since \(x \geq 0\)).
1. **Sketch the curves**: The line \(y = x\) is a straight line through the origin with a slope of 1. The curve \(y = x^3\) is a cubic function that also passes through the origin and grows more slowly than the line for small values of x.
2. **Find their points of intersection**: We set the equations equal to find where they intersect:
\[ x = x^3 \implies x^3 – x = 0 \implies x(x^2 – 1) = 0. \]
This gives us \(x = 0\) and \(x = 1\). Thus, the intersection points are (0, 0) and (1, 1).
3. **Determine the limits of integration**: The region D can be described by horizontal strips (dy) or vertical strips (dx). We choose dx as our variable of integration. For a fixed \(x\), the limits for \(y\) vary from the cubic curve up to the line:
\(y_{bottom} = x^3\) to \(y_{top} = x\).
4. **Set up the double integral**:
\[ \int_{0}^{1} \int_{x^3}^{x} (x^2 + 6y) \, dy \, dx. \]
5. **Evaluate the inner integral**:
\[ \int_{x^3}^{x} (x^2 + 6y) \, dy = \left[ x^2 y + 3y^2 \right]_{y=x^3}^{y=x} = (x^2(x) + 3(x)^2) – (x^2(x^3) + 3(x^3)^2). \]
Simplifying gives:
\[x^3 + 3x^2 – (x^5 + 3x^6) = x^3 + 3x^2 – x^5 – 3x^6. \]
6. **Now evaluate the outer integral**:
\[ \int_{0}^{1} (x^3 + 3x^2 – x^5 – 3x^6) \, dx = \left[ \frac{x^4}{4} + x^3 – \frac{x^6}{6} – \frac{3x^7}{7} \right]_{0}^{1}. \]
7. **Substituting the limits**:
At \(x=1\): \(\frac{1}{4} + 1 – \frac{1}{6} – \frac{3}{7} = \frac{42}{84} + \frac{84}{84} – \frac{14}{84} – \frac{36}{84} = \frac{76}{84} = \frac{38}{42} = \frac{19}{21}. \)
8. Therefore, the value of the double integral is \(\frac{19}{21}\).